Let $f(x) = 7x^{2}+7x-4$. Where does this function intersect the x-axis (i.e. what are the roots or zeroes of $f(x)$ )?
Solution: The function intersects the x-axis when $f(x) = 0$ , so you need to solve the equation: $7x^{2}+7x-4 = 0$ Use the quadratic formula to solve $ax^2 + bx + c = 0$ $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ $a = 7, b = 7, c = -4$ $ x = \dfrac{-7 \pm \sqrt{7^{2} - 4 \cdot 7 \cdot -4}}{2 \cdot 7}$ $ x = \dfrac{-7 \pm \sqrt{161}}{14}$ $ x = \dfrac{-7 \pm \sqrt{161}}{14}$ $x =\dfrac{-7 \pm \sqrt{161}}{14}$